Integral of $$$\ln\left(\sqrt{3} x\right)$$$

The calculator will find the integral/antiderivative of $$$\ln\left(\sqrt{3} x\right)$$$, with steps shown.

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Find $$$\int \ln\left(\sqrt{3} x\right)\, dx$$$.

Solution

Let $$$u=\sqrt{3} x$$$.

Then $$$du=\left(\sqrt{3} x\right)^{\prime }dx = \sqrt{3} dx$$$ (steps can be seen »), and we have that $$$dx = \frac{\sqrt{3} du}{3}$$$.

The integral can be rewritten as

$${\color{red}{\int{\ln{\left(\sqrt{3} x \right)} d x}}} = {\color{red}{\int{\frac{\sqrt{3} \ln{\left(u \right)}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\sqrt{3}}{3}$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sqrt{3} \ln{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\sqrt{3} \int{\ln{\left(u \right)} d u}}{3}\right)}}$$

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$.

Let $$$\operatorname{\omega}=\ln{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{d\omega}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

The integral can be rewritten as

$$\frac{\sqrt{3} {\color{red}{\int{\ln{\left(u \right)} d u}}}}{3}=\frac{\sqrt{3} {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{3}=\frac{\sqrt{3} {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{3}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$\frac{\sqrt{3} \left(u \ln{\left(u \right)} - {\color{red}{\int{1 d u}}}\right)}{3} = \frac{\sqrt{3} \left(u \ln{\left(u \right)} - {\color{red}{u}}\right)}{3}$$

Recall that $$$u=\sqrt{3} x$$$:

$$\frac{\sqrt{3} \left(- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}\right)}{3} = \frac{\sqrt{3} \left(- {\color{red}{\sqrt{3} x}} + {\color{red}{\sqrt{3} x}} \ln{\left({\color{red}{\sqrt{3} x}} \right)}\right)}{3}$$

Therefore,

$$\int{\ln{\left(\sqrt{3} x \right)} d x} = \frac{\sqrt{3} \left(\sqrt{3} x \ln{\left(\sqrt{3} x \right)} - \sqrt{3} x\right)}{3}$$

Simplify:

$$\int{\ln{\left(\sqrt{3} x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \frac{\ln{\left(3 \right)}}{2}\right)$$

Add the constant of integration:

$$\int{\ln{\left(\sqrt{3} x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \frac{\ln{\left(3 \right)}}{2}\right)+C$$

Answer

$$$\int \ln\left(\sqrt{3} x\right)\, dx = x \left(\ln\left(x\right) - 1 + \frac{\ln\left(3\right)}{2}\right) + C$$$A


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